\(\text{∘ Ans}\)
\(\downarrow\)
\(A=\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-...-\dfrac{1}{6}-\dfrac{1}{2}\)
`=`\(\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)\)
`=`\(\dfrac{8}{9}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\right)\)
`=`\(\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
`=`\(\dfrac{8}{9}-\left[1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-...-\left(\dfrac{1}{8}-\dfrac{1}{8}\right)-\dfrac{1}{9}\right]\)
`=`\(\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)\)
`=`\(\dfrac{8}{9}-\dfrac{8}{9}=0\)
Vậy, ` A = 0.`
\(A=\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)=\)
\(A=\dfrac{8}{9}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{7.8}+\dfrac{1}{8.9}\right)=\)
\(A=\dfrac{8}{9}-\left(\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{9-8}{8.9}\right)\)
\(A=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
\(A=\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)=0\)
\(...A=\dfrac{8}{9}-\dfrac{1}{8.9}-\dfrac{1}{7.8}-\dfrac{1}{6.7}-...\dfrac{1}{2.3}-\dfrac{1}{2.1}\)
\(A=\dfrac{8}{9}+\dfrac{1}{9}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{6}-...+\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{1}=\left(\dfrac{8}{9}+\dfrac{1}{9}\right)-1=0\)
\(\dfrac{8}{9}\)-(\(\dfrac{1}{2}\)+\(\dfrac{1}{6}\)+...+\(\dfrac{1}{56}\)+\(\dfrac{1}{72}\))
=\(\dfrac{8}{9}\)-(\(\dfrac{1}{1\cdot2}\)+\(\dfrac{1}{2\cdot3}\)+...+\(\dfrac{1}{7\cdot8}\)+\(\dfrac{1}{8\cdot9}\))
=\(\dfrac{8}{9}\)-(1-\(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+...+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\))
=\(\dfrac{8}{9}\)-(1-\(\dfrac{1}{9}\))
=\(\dfrac{8}{9}\)-\(\dfrac{8}{9}\)=0
